# Convergence in distribution (central limit theorem)

If $$X_1, ... , X_n$$ be iid exponential with mean $$1/\lambda$$. Let $$S_n = X_1 + ... + X_n$$.

a) Show that $$S_n$$ is $$\Gamma(n, 1/\lambda)$$.

Each $$X_i$$ is $$\Gamma(1, 1/\lambda)$$ by the definition of an exponential distribution. We can look at the mgf of $$S_n$$.

$$M(t) = E(e^{t \sum X_i}) = E(e^{tX_1})...E(e^{tX_n}) = (\frac{1}{1-t/\lambda})...(\frac{1}{1-t/\lambda}) = (\frac{1}{1-t/\lambda})^n,$$

which is the mgf of $$\Gamma(n, 1/\lambda)$$. So $$S_n \sim \Gamma(n, 1/\lambda)$$.

b) Show that $$\frac{(S_n - n/\lambda)}{\sqrt{n}}$$ converges in distribution and identify the asymptotic distribution.

By the Central Limit Theorem, we know that $$\frac{(S_n - n\mu)}{\sqrt{n}(1/\lambda^2)} = \frac{(S_n/n - \mu)}{(1/\lambda^2)(1/\sqrt{n})}$$ converges in distribution to $$N(0,1)$$.

We see that $$\frac{(S_n/n - 1/\lambda)}{(1/\lambda^2)(1/\sqrt{n})} = \frac{(S_n/\sqrt{n} - \sqrt{n}/\lambda)}{(1/\lambda^2)}(\frac{\sqrt{n}}{\sqrt{n}}) = \frac{S_n - n/\lambda}{(1/\lambda^2)(\sqrt{n})}$$

So, $$\frac{S_n - n/\lambda}{(1/\lambda^2)(\sqrt{n})} \xrightarrow{D} N(0,1)$$ implies that $$\frac{S_n - n/\lambda}{(\sqrt{n})} = (1/\lambda^2)\frac{S_n - n/\lambda}{(1/\lambda^2)(\sqrt{n})} \xrightarrow{D} N(0, 1/\lambda^4).$$

c) Show that $$\sqrt{n}(\sqrt{S_n/n} - \sqrt{1/\lambda})$$ converges in distribution and identify the asymptotic distribution.

There is a theorem in our textbook (Introduction to Mathematical Statistics by Hogg) which states,

Let $$\{X_n\}$$ be a sequence of random variables such that $$\sqrt{n}(X_n - \theta) \xrightarrow{D} N(0, \sigma^2)$$. Suppose the function $$g(x)$$ is differentiable at $$\theta$$ and $$g'(\theta) \not= 0$$. Then $$\sqrt{n}(g(X_n) - g(\theta)) \xrightarrow{D} N(0, \sigma^2(g'(\theta)^2)$$.

From part b), we see that $$\frac{\sqrt{n}(S_n/n - 1/\lambda)}{(1/\lambda^2)} \rightarrow N(0,1).$$

So $$\sqrt{n}(S_n/n - 1/\lambda) \rightarrow N(0, 1/\lambda^4)$$.

Now we apply the theorem. We have $$g(x)=\sqrt{x}$$. So $$g'(x)=1/(2\sqrt{x})$$, and $$g'(\frac{1}{\lambda}) = \frac{\sqrt{\lambda}}{2}$$.

So by the theorem,

$$\sqrt{n}(\sqrt{S_n/n} - \sqrt{1/\lambda}) \rightarrow N(0, (\frac{\sqrt{\lambda}}{2\lambda^4})).$$

Do you think my answers are correct?

• This is a little nit-picky, but in (b) you need to justify the applicability of the CLT. To see why this might be necessary, re-work your answers replacing the exponential distribution with a Cauchy distribution. – whuber Jan 9 '14 at 22:02
• At the end of c), shouldn't you square the derivative of $g$? – Alecos Papadopoulos Jan 9 '14 at 22:07
• There are a few more algebraic mistakes in (c), too. Consider running a simulation for a tractable $\lambda$ (such as $\lambda=3$) and increasing values of $n$ (to establish near-convergence of the results) as a check of your formulas. It takes only a few minutes. – whuber Jan 9 '14 at 22:34
• @whuber Thanks a lot. In our textbook, it says "Let $X_1, ... , X_n$ denote the observation that has mean $\mu$ and positive variance $\sigma^2$. Then the random variable $Y_n = (\sum^n_{i=1} X_i - n \mu)/\sqrt{n}\sigma = \sqrt{n}(\bar{X_n} - \mu)\sigma$ converges in distribution to a random variable which has a normal distribution with mean zero and variance 1." Since, in this case, the $X_i$ are iid and $\sigma^2 = n/\lambda^2 > 0$, we can use the CLT, right? – Artus Jan 10 '14 at 12:16
• Yes, that's correct: your comment supplies the justification for applying the CLT. – whuber Jan 10 '14 at 15:03

$$\sqrt{n}\left(\sqrt{S_n/n} - \sqrt{1/\lambda}\right) \rightarrow N\left(0, \frac{1}{4\lambda^3}\right)$$